0.000 001 506 198 729 391 92 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 001 506 198 729 391 92(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.000 001 506 198 729 391 92(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 001 506 198 729 391 92.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 001 506 198 729 391 92 × 2 = 0 + 0.000 003 012 397 458 783 84;
  • 2) 0.000 003 012 397 458 783 84 × 2 = 0 + 0.000 006 024 794 917 567 68;
  • 3) 0.000 006 024 794 917 567 68 × 2 = 0 + 0.000 012 049 589 835 135 36;
  • 4) 0.000 012 049 589 835 135 36 × 2 = 0 + 0.000 024 099 179 670 270 72;
  • 5) 0.000 024 099 179 670 270 72 × 2 = 0 + 0.000 048 198 359 340 541 44;
  • 6) 0.000 048 198 359 340 541 44 × 2 = 0 + 0.000 096 396 718 681 082 88;
  • 7) 0.000 096 396 718 681 082 88 × 2 = 0 + 0.000 192 793 437 362 165 76;
  • 8) 0.000 192 793 437 362 165 76 × 2 = 0 + 0.000 385 586 874 724 331 52;
  • 9) 0.000 385 586 874 724 331 52 × 2 = 0 + 0.000 771 173 749 448 663 04;
  • 10) 0.000 771 173 749 448 663 04 × 2 = 0 + 0.001 542 347 498 897 326 08;
  • 11) 0.001 542 347 498 897 326 08 × 2 = 0 + 0.003 084 694 997 794 652 16;
  • 12) 0.003 084 694 997 794 652 16 × 2 = 0 + 0.006 169 389 995 589 304 32;
  • 13) 0.006 169 389 995 589 304 32 × 2 = 0 + 0.012 338 779 991 178 608 64;
  • 14) 0.012 338 779 991 178 608 64 × 2 = 0 + 0.024 677 559 982 357 217 28;
  • 15) 0.024 677 559 982 357 217 28 × 2 = 0 + 0.049 355 119 964 714 434 56;
  • 16) 0.049 355 119 964 714 434 56 × 2 = 0 + 0.098 710 239 929 428 869 12;
  • 17) 0.098 710 239 929 428 869 12 × 2 = 0 + 0.197 420 479 858 857 738 24;
  • 18) 0.197 420 479 858 857 738 24 × 2 = 0 + 0.394 840 959 717 715 476 48;
  • 19) 0.394 840 959 717 715 476 48 × 2 = 0 + 0.789 681 919 435 430 952 96;
  • 20) 0.789 681 919 435 430 952 96 × 2 = 1 + 0.579 363 838 870 861 905 92;
  • 21) 0.579 363 838 870 861 905 92 × 2 = 1 + 0.158 727 677 741 723 811 84;
  • 22) 0.158 727 677 741 723 811 84 × 2 = 0 + 0.317 455 355 483 447 623 68;
  • 23) 0.317 455 355 483 447 623 68 × 2 = 0 + 0.634 910 710 966 895 247 36;
  • 24) 0.634 910 710 966 895 247 36 × 2 = 1 + 0.269 821 421 933 790 494 72;
  • 25) 0.269 821 421 933 790 494 72 × 2 = 0 + 0.539 642 843 867 580 989 44;
  • 26) 0.539 642 843 867 580 989 44 × 2 = 1 + 0.079 285 687 735 161 978 88;
  • 27) 0.079 285 687 735 161 978 88 × 2 = 0 + 0.158 571 375 470 323 957 76;
  • 28) 0.158 571 375 470 323 957 76 × 2 = 0 + 0.317 142 750 940 647 915 52;
  • 29) 0.317 142 750 940 647 915 52 × 2 = 0 + 0.634 285 501 881 295 831 04;
  • 30) 0.634 285 501 881 295 831 04 × 2 = 1 + 0.268 571 003 762 591 662 08;
  • 31) 0.268 571 003 762 591 662 08 × 2 = 0 + 0.537 142 007 525 183 324 16;
  • 32) 0.537 142 007 525 183 324 16 × 2 = 1 + 0.074 284 015 050 366 648 32;
  • 33) 0.074 284 015 050 366 648 32 × 2 = 0 + 0.148 568 030 100 733 296 64;
  • 34) 0.148 568 030 100 733 296 64 × 2 = 0 + 0.297 136 060 201 466 593 28;
  • 35) 0.297 136 060 201 466 593 28 × 2 = 0 + 0.594 272 120 402 933 186 56;
  • 36) 0.594 272 120 402 933 186 56 × 2 = 1 + 0.188 544 240 805 866 373 12;
  • 37) 0.188 544 240 805 866 373 12 × 2 = 0 + 0.377 088 481 611 732 746 24;
  • 38) 0.377 088 481 611 732 746 24 × 2 = 0 + 0.754 176 963 223 465 492 48;
  • 39) 0.754 176 963 223 465 492 48 × 2 = 1 + 0.508 353 926 446 930 984 96;
  • 40) 0.508 353 926 446 930 984 96 × 2 = 1 + 0.016 707 852 893 861 969 92;
  • 41) 0.016 707 852 893 861 969 92 × 2 = 0 + 0.033 415 705 787 723 939 84;
  • 42) 0.033 415 705 787 723 939 84 × 2 = 0 + 0.066 831 411 575 447 879 68;
  • 43) 0.066 831 411 575 447 879 68 × 2 = 0 + 0.133 662 823 150 895 759 36;
  • 44) 0.133 662 823 150 895 759 36 × 2 = 0 + 0.267 325 646 301 791 518 72;
  • 45) 0.267 325 646 301 791 518 72 × 2 = 0 + 0.534 651 292 603 583 037 44;
  • 46) 0.534 651 292 603 583 037 44 × 2 = 1 + 0.069 302 585 207 166 074 88;
  • 47) 0.069 302 585 207 166 074 88 × 2 = 0 + 0.138 605 170 414 332 149 76;
  • 48) 0.138 605 170 414 332 149 76 × 2 = 0 + 0.277 210 340 828 664 299 52;
  • 49) 0.277 210 340 828 664 299 52 × 2 = 0 + 0.554 420 681 657 328 599 04;
  • 50) 0.554 420 681 657 328 599 04 × 2 = 1 + 0.108 841 363 314 657 198 08;
  • 51) 0.108 841 363 314 657 198 08 × 2 = 0 + 0.217 682 726 629 314 396 16;
  • 52) 0.217 682 726 629 314 396 16 × 2 = 0 + 0.435 365 453 258 628 792 32;
  • 53) 0.435 365 453 258 628 792 32 × 2 = 0 + 0.870 730 906 517 257 584 64;
  • 54) 0.870 730 906 517 257 584 64 × 2 = 1 + 0.741 461 813 034 515 169 28;
  • 55) 0.741 461 813 034 515 169 28 × 2 = 1 + 0.482 923 626 069 030 338 56;
  • 56) 0.482 923 626 069 030 338 56 × 2 = 0 + 0.965 847 252 138 060 677 12;
  • 57) 0.965 847 252 138 060 677 12 × 2 = 1 + 0.931 694 504 276 121 354 24;
  • 58) 0.931 694 504 276 121 354 24 × 2 = 1 + 0.863 389 008 552 242 708 48;
  • 59) 0.863 389 008 552 242 708 48 × 2 = 1 + 0.726 778 017 104 485 416 96;
  • 60) 0.726 778 017 104 485 416 96 × 2 = 1 + 0.453 556 034 208 970 833 92;
  • 61) 0.453 556 034 208 970 833 92 × 2 = 0 + 0.907 112 068 417 941 667 84;
  • 62) 0.907 112 068 417 941 667 84 × 2 = 1 + 0.814 224 136 835 883 335 68;
  • 63) 0.814 224 136 835 883 335 68 × 2 = 1 + 0.628 448 273 671 766 671 36;
  • 64) 0.628 448 273 671 766 671 36 × 2 = 1 + 0.256 896 547 343 533 342 72;
  • 65) 0.256 896 547 343 533 342 72 × 2 = 0 + 0.513 793 094 687 066 685 44;
  • 66) 0.513 793 094 687 066 685 44 × 2 = 1 + 0.027 586 189 374 133 370 88;
  • 67) 0.027 586 189 374 133 370 88 × 2 = 0 + 0.055 172 378 748 266 741 76;
  • 68) 0.055 172 378 748 266 741 76 × 2 = 0 + 0.110 344 757 496 533 483 52;
  • 69) 0.110 344 757 496 533 483 52 × 2 = 0 + 0.220 689 514 993 066 967 04;
  • 70) 0.220 689 514 993 066 967 04 × 2 = 0 + 0.441 379 029 986 133 934 08;
  • 71) 0.441 379 029 986 133 934 08 × 2 = 0 + 0.882 758 059 972 267 868 16;
  • 72) 0.882 758 059 972 267 868 16 × 2 = 1 + 0.765 516 119 944 535 736 32;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 001 506 198 729 391 92(10) =

0.0000 0000 0000 0000 0001 1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0111 0100 0001(2)

5. Positive number before normalization:

0.000 001 506 198 729 391 92(10) =

0.0000 0000 0000 0000 0001 1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0111 0100 0001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right, so that only one non zero digit remains to the left of it:


0.000 001 506 198 729 391 92(10) =

0.0000 0000 0000 0000 0001 1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0111 0100 0001(2) =

0.0000 0000 0000 0000 0001 1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0111 0100 0001(2) × 20 =

1.1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0111 0100 0001(2) × 2-20


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -20

Mantissa (not normalized):
1.1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0111 0100 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-20 + 2(11-1) - 1 =

(-20 + 1 023)(10) =

1 003(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 003 ÷ 2 = 501 + 1;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1003(10) =

011 1110 1011(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0111 0100 0001 =

1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0111 0100 0001


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1110 1011

Mantissa (52 bits) =
1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0111 0100 0001


Decimal number 0.000 001 506 198 729 391 92 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1011 - 1001 0100 0101 0001 0011 0000 0100 0100 0110 1111 0111 0100 0001

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100